I was playing trying to count how many union-closed families $\cal{F} \subseteq \cal{P}(\{a, b, c\}) \setminus \emptyset$ are there.
I know there are $61$ such families (see for example this OEIS sequence), but I want to count them using generating functions.
To do that I have defined for any $A \in \cal{P}(\{a, b, c\})$, $A \neq \emptyset$:
$$x_A = \begin{cases} 1, & \text{if $A \in \cal{F}$} \\ 0, & \text{if $A \notin \cal{F}$} \end{cases}$$
Since for example, when $x_{\{a\}}=1$ and $x_{\{b\}}=1$ it must be $x_{\{a,b\}}=1$, we can count how many solutions the following system has:
$$\begin{cases} x_{\{a\}}+x_{\{b\}}-x_{\{a,b\}} \neq 2 \\ x_{\{a\}}+x_{\{c\}}-x_{\{a,c\}} \neq 2 \\ x_{\{b\}}+x_{\{c\}}-x_{\{b,c\}} \neq 2 \\ x_{\{a,b\}}+x_{\{b,c\}}-x_{\{a,b,c\}} \neq 2 \\ x_{\{a,b\}}+x_{\{a,c\}}-x_{\{a,b,c\}} \neq 2 \\ x_{\{a,c\}}+x_{\{b,c\}}-x_{\{a,b,c\}} \neq 2 \\ x_{\{a\}}+x_{\{b,c\}}-x_{\{a,b,c\}} \neq 2 \\ x_{\{b\}}+x_{\{a,c\}}-x_{\{a,b,c\}} \neq 2 \\ x_{\{c\}}+x_{\{a,b\}}-x_{\{a,b,c\}} \neq 2 \\ \end{cases}$$
Then, using the method explained here we get the following generating function:
$$g(z_1,\ldots,z_7)=(1+z_1z_2z_7)(1+z_1z_3z_8)(1+z_2z_3z_9)(1+z_4z_5z_9/z_1)(1+z_5z_6z_8/z_2)(1+z_4z_6z_7/z_3)(1+1/z_4z_5z_6z_7z_8z_9)$$
and expanding e.g. with Wolfram Alpha and a little patience one can count $61$ terms where no $z_i$ $i=1,\ldots,7$ has a power equal to $2$.
Since it would be nice to extend the method for more than three elements, my question is:
Is there a more clever way of getting the coefficients of the above function (or the sum of those with no $z_i^2$), without expanding or differentiating, so something with a lower computational complexity, for example proportional to the number $2^k-1$ ($k$ being the number of elements, $k = 3$ in the showed case) of the $x_A$ variables?
